Modeling salt and water with differential equationsA 5-gallon bucket is full of pure water. Suppose we begin dumping salt into the bucket at a rate of 1/4 pounds per minute. Also, we open the spigot so that 1/2 gallons per minute leaves the bucket, and we add pure water to keep the bucket full. If the salt water solution is always well mixed, what is the amount of salt in the bucket after&lt;br?(a) 1 minute?(b) 10 minutes?(c) 10 minutes?(d) 1000 minutes?(e) a very, very long time?My question is about the model. I let x be the amount of salt and t the minutes passed. Then d x d t = 1 4 − 1 10 xSome algebra gives me d x d t = 5 − 2 x 20 and I integrate ∫ 20 5 − 2 x d x = ∫ d tto obtain − 10 ln ⁡ | 5 − 2 x | = t + C. After some algebra, and keeping in mind that x ( 0 ) = 0, I get | 5 − 2 x | = 5 e − t 10 I have two sets of solutions: x = 5 2 ( 1 − e − t 10 ) and x = 5 2 ( 1 + e − t 10 ) At this point I'm confused about what to do. Either model provides "sensible" answers, in that the amount of salt I have is positive, as we might expect. So which one do I use?

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Modeling salt and water with differential equationsA 5-gallon bucket is full of pure water. Suppose we begin dumping salt into the bucket at a rate of 1/4 pounds per minute. Also, we open the spigot so that 1/2 gallons per minute leaves the bucket, and we add pure water to keep the bucket full. If the salt water solution is always well mixed, what is the amount of salt in the bucket after&amp;lt;br?(a) 1 minute?(b) 10 minutes?(c) 10 minutes?(d) 1000 minutes?(e) a very, very long time?My question is about the model. I let x be the amount of salt and t the minutes passed. Then            d      x              d      t        =      1    4    −      1    10    xSome algebra gives me            d      x              d      t        =            5      −      2      x        20  and I integrate  ∫      20          5      −      2      x        d  x  =  ∫    d  tto obtain   −  10  ln  ⁡      |    5  −  2  x      |    =  t  +  C. After some algebra, and keeping in mind that   x  (  0  )  =  0, I get      |    5  −  2  x      |    =  5      e          −              t        10            I have two sets of solutions:   x  =      5    2        (    1    −          e              −                  t          10                      )   and   x  =      5    2        (    1    +          e              −                  t          10                      )  At this point I&#039;m confused about what to do. Either model provides &quot;sensible&quot; answers, in that the amount of salt I have is positive, as we might expect. So which one do I use?

Samuel Hooper · Accepted Answer

Step 1Keeping in mind that   x  (  0  )  =  0 you might wish to rethink the statement that       5    2        (    1    +          e              −        t                  /                10              )   hwich has   x  (  0  )  =  5 is &quot;sensible.&quot;Step 2As an aside, you also may wish to recheck your units: A pound of salt is not equivalent to a gallon of water.

klesstilne1 · Accepted Answer

Step 1So before anything else, your second solution has   x  (  0  )  =  5, so that can&#039;t be right.As for what really happened, you got the general implicit solution:  −  10  ln  ⁡      |    5  −  2  x      |    =  t  +  CSince   x  (  0  )  =  0, the left side starts out at   −  10  ln  ⁡  (  5  ), with the positive logarithm. It will not cross over into the other one; if that ever happened, from continuity your division by   5  −  2  x would become a division by 0 which would be invalid. So you&#039;ll have the positive logarithm the entire time. Accordingly:  −  10  ln  ⁡  (  5  −  2  x  )  =  t  +  C    −  10  ln  ⁡  (  5  )  =  C    −  10  ln  ⁡  (  5  −  2  x  )  =  t  −  10  ln  ⁡  (  5  )    10  ln  ⁡  (  5  −  2  x  )  =  10  ln  ⁡  (  5  )  −  t    10  ln  ⁡  (  (  5  −  2  x  )      /    5  )  =  −  t    (  5  −  2  x  )      /    5  =      e          −      t              /            10       etc.The cleanest explanation, which also avoids the issue of treating differentials as separate variables, is to write      20          5      −      2      y                  d      y              d      s        =  1and then integrate both sides with respect to s:      ∫    0    t        20          5      −      2      y                  d      y              d      s        d  s  =      ∫    0    t    d  sthen change variables on the left side:      ∫          x      (      0      )              x      (      t      )            20          5      −      2      y        d  y  =      ∫    0    t    d  sStep 2Then change variables again, and twiddle some minus signs:      ∫          5      −      2      x      (      t      )              5      −      2      x      (      0      )            10    u    d  u  =      ∫    0    t    d  sNow the integral on the left side is over a positive range of u, since   5  −  2  x  (  0  )  =  5  &amp;gt;  0, so we wind up with a positive logarithm on the left side.

A 5-gallon bucket is full of pure water. Suppose we begin dumping salt into the bucket at a rate of 1/4 pounds per minute. Also, we open the spigot so that 1/2 gallons per minute leaves the bucket, and we add pure water to keep the bucket full.

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